Equation of an Off-Center Circle
This is a standard example that comes up a lot. Circles are easy to describe,
unless the origin is on the rim of the circle. We’ll calculate the equation in polar
coordinates of a circle with center (a, 0) and radius (2a, 0). You should expect
to repeat this calculation a few times in this class and then memorize it for
multivariable calculus, where you’ll need it often.
(a,0)
Figure 1: Off center circle through (0, 0).
In rectangular coordinates, the equation of this circle is:
(x − a)2 + y 2 = a2 .
We could plug in x = r cos θ, y = sin θ to convert to polar coordinates, but
there’s a faster way. We start by expanding and simplifying:
(x − a)2 + y 2
=
a2
x2 − 2ax + a2 + y 2
=
a2
x2 − 2ax + y 2 =
(x2 + y 2 ) − 2ax =
r2 − 2ar cos θ =
0
0
0
2
=
2ar cos θ
=⇒ r
=
2a cos θ
r
(or r = 0).
We used the facts that x2 + y 2 = r2 and x = r cos θ to conclude that there
were two values of r that satisfy this equation; r = 2a cos θ and r = 0. These
are the equations describing r in terms of θ that describe this circle in polar
coordinates.
In order to use the equation r = 2a cos θ, we need to figure out the appro­
priate range of values for θ. By looking at the graph we see that − π2 < θ < π2 .
Our final equation is:
r=0
or
r = 2a cos θ,
1
−
π
π
<θ< .
2
2
y
r
θ
(a,0)
x
Figure 2: Off center circle in polar coordinates.
To check our work, let’s find some points on this curve:
• At θ = 0, r = 2a and so x = 2a and y = 0.
√
• At θ = π4 , r = 2a cos π4 = a 2. Hence x = a and y = a.
2
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18.01SC Single Variable Calculus��
Fall 2010 ��
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